Question

${lim}_{h\to 0}\left\{\frac{1}{h\sqrt[3]{38+h}}-\frac{1}{2h}\right\}$ is equal to ?

• A. $\frac{1}{12}$
• B. $-\frac{4}{3}$
• C. $-\frac{16}{3}$
• D. $-\frac{1}{48}$

Answer 1

The correct option is D $-\frac{1}{48}$

${lim}_{h\to 0}(\frac{1}{h\sqrt[3]{38+h}}-\frac{1}{2h})={lim}_{h\to 0}\frac{2-\sqrt[3]{38+h}}{2h\sqrt[3]{38+h}}={lim}_{h\to 0}\frac{8-(8+h)}{2h\sqrt[3]{38+h}(8^{2/3}+8^{1/3}{(8+h)}^{1/3}+{(8+h)}^{2/3})}=-\frac{1}{48}$
Hence, option 'D' is correct.